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A 5.0-kg box is pushed across the floor with a force of 32 N. What is the frictional force on the box if the box's acceleration is 3.6 m/s2?


Sagot :

luana
[tex]From\ 2nd\ Newton\ Law\ resultuant\ force: \\\\\ F_R=m*a=5kg*3,6\frac{m}{s^2}=18N\\\\ F=32N\\\\ F_f- friction\ force\\\\ F_R=F-F_f\\\\ F_f=F-F_R=32N-18N=14N\\\\Friction\ force\ is\ equal\ to\ 14\ N.[/tex]
AL2006
In order to accelerate 5.0 kg of mass at the rate of 3.6 m/s², the force required is

F = m a = (5) (3.6) = 18 newtons.

The force applied is 32 newtons.  So 18 of them are going to accelerate the box,
and friction is eating up the other 14 newtons by acting in the opposite direction.
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